class ex { public static void main(String a[]) { int i=1; System.out.println(i++ + i++); System.out.println(((i++) + (i++))); System.out.println((i++) + (i++)); } } The output is 3, 7, 11. First output is clear. Please clarify last two outputs. Thanx

Monika Pasricha
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Joined: Jan 08, 2001
Posts: 19

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I got the answer.. The reason is same as for the first output.

The value of i after first SOP statement becomes 2.In second SOP statement as precedence is given to the (),inside SOP statement (i++) sets i =3 and then another (i++) sets it to 4.therefore 3+4=7.In third SOP the same thing happens as in second SOP, hence 5+6 = 11 simple isn't it!!

The split up is as follows System.out.println(1 + 2) i=3 System.out.println(((3) + (4))) i=5 System.out.println((5) + (6)) i=7 } } The output is 3, 7, 11. Thanx uma

initial value of i=1; Consider step 1: Let me write i++ + i++ as a + b where a=i++ before the binary + operator and b=i++ after the binary + operator. Since a post increment is being done,the value of a is 1, value of i is i+1 i.e 2. Now the current value of i is 2, so the value of b is i which is 2 and the value of i is incremented to 3. So final result after step 1 is: a=1, b=2, i=3 So the results is printed as (a+b)=(1+2)=3 Write the expression in step 2, again in terms of a and b. Then the expression would be (a+b). Again by appliying the same procdeure as above, the value of a is current value of i which is 3 , the value of i after this post increment is 4, the value of b is current value of i, which is 4 and the value of i after this increment is 5. So the final values are a=3,b=4 and i=5, giing a result of a+b=3+4=7. Similarly, by applying the same procedure to 3,value of a is current value of i which is 5, i is incremented to 6, value of b is current value of i which is 6, i is incremented to 7. Final result is a+b=5+6 = 11. So, the algorithm can be summarised as: 1)Let a=i++ and b=i++ 2) The evaluation goes as follows, a=i, i=i+1, b=i, i=i+1,a+b(from left to right).Note that the paranthesis is steps 2, and 3 are redundant and are of no significance in this case! Hope this helps!

Originally posted by Monika Pasricha: [B]class ex { public static void main(String a[]) { int i=1; System.out.println(i++ + i++);

i ++ = 1 (means the value of i and pass ++ i.e 1 and pass 2) i ++ = 2 (means the value of i and pass ++ i.e. 2 and pass 3) so te result of first statement is 3 (i is 2 here)

System.out.println(((i++) + (i++))); (i++) = 3 (means the value of i and ++ (2++) = (3)) (i++) = 4 (means the value of i and ++ (3++) = (4)) so the result here is 3 + 4 = 7 ( i is 4 here) System.out.println((i++) + (i++)); (i++) = 5 (means the value of i and ++ (4++) = (5)) (i++) = 6 (means the value of i and ++ (5++) = (6)) so the result here is 5 + 6 = 11 } } The output is 3, 7, 11.

the simplest way to explain is this for post increment operations: The increment works as follows: The first one: sop(1(2) + 2(3)) total is 3 but the value of i is 3 second one: sop(3(4) + 4(5)) total is 7 and the value of i is 7 third one: sop(5(6) + 6(7)) total is 11 and the value of i 7 Trust this sequencing should help venkat